Finding a reduction formula

Question

Find a reduction formula for $$I_n= \int x \cos^n{x} dx.$$ I have used integration by parts as $$u=x, dv=\cos^n{x}dx,$$ then $$I_n=x\int\cos^n{x}dx-\int\left(\int\cos^n{x}\right)dx,$$ where $$I^*_n=\int\cos^n{x}dx$$ is the well-known reduction formula, then $$I_n=xI^*_n-\int I^*_n dx.$$ Is this correct, or is there another answer?

Answer 1

For $n \geqslant 2$, \begin{align*} I_n &= \int x \cos^n x \,\mathrm{d}x = \int x \cos^{n - 1} x\,\mathrm{d}(\sin x)\\ & = x \sin x \cos^{n - 1} x - \int \sin x \,\mathrm{d}(x \cos^{n - 1} x)\\ &= x \sin x \cos^{n - 1} x - \left(\int \sin x \cos^{n - 1} x\,\mathrm{d}x + \int x \sin x \cdot (n - 1) \cos^{n - 2} x (-\sin x) \,\mathrm{d}x\right)\\ &= x \sin x \cos^{n - 1} x + \int \cos^{n - 1} x\,\mathrm{d}(\cos x) + (n - 1) \int x \sin^2 x \cos^{n - 2} x \,\mathrm{d}x\\ &= x \sin x \cos^{n - 1} x + \frac{1}{n} \cos^n x + (n - 1) \int x (1 - \cos^2 x) \cos^{n - 2} x \,\mathrm{d}x\\ &= x \sin x \cos^{n - 1} x + \frac{1}{n} \cos^n x + (n - 1) \int x \cos^{n - 2} x \,\mathrm{d}x - (n - 1) \int x \cos^n x \,\mathrm{d}x\\ &= x \sin x \cos^{n - 1} x + \frac{1}{n} \cos^n x + (n - 1)I_{n - 2} - (n - 1)I_n, \end{align*} thus$$ I_n = \frac{1}{n} x \sin x \cos^{n - 1} x + \frac{1}{n^2} \cos^n x + \frac{n - 1}{n} I_{n - 2}. $$